<em>Gasoline</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>
<em>acetone</em><em> </em><em>in</em><em> </em><em>nail</em><em> </em><em>polished</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>
<em>salt</em><em> </em><em>in</em><em> </em><em>alcohol</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>
<em>oil</em><em> </em><em>in</em><em> </em><em>vinegar</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>
<em>tawas</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>soluble</em><em>.</em><em>.</em><em>.</em>
<em>Sorry</em><em> </em><em>if</em><em> </em><em>i</em><em> </em><em>am</em><em> </em><em>incorrect</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.
Explanation:
Temperature of Solid
Melting temperature of Solid 
Temperature of liquid 
Specific heats of solid ethanol = 0.97 J/gK
Specific heats of liquid ethanol = 2.3 J/gK
Heat required to melt the the 25 g solid
at 159 K
= 159 K - 138 K = 21 K

Heat required to melt and raise the temperature of
upto 223 K
= 223 K - 159 K = 64 K

Total heat to convert solid ethanol to liquid ethanol at given temperature :
(1kJ=1000J)
Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.
1. Common knowledge : Go to the right of periodic table, the atomic radius is decreasing
2. Flurine has 9 protons and lithium has 3 protons. you know that the electron is attracted with the centre of the atom, that's why more proton, more 'energy' that attract to the centre and that's why it make the shape of the atom is smaller
Answer:
mm = 1043.33 g/mol
Explanation:
osmotic pressure (π):
∴ π = 17.8 torr = 0.0234 atm
∴ Cb: solute concentration
∴ T = 25°C = 298 K
∴ R = 0.082 atm.L/K.mol
⇒ Cb = π/RT
⇒ Cb = (0.0234 atm)/((0.082 atm.L/K.mol)(298 K))
⇒ Cb = 9.585 E-4 mol/L
molar mass (mm):
⇒ mm = (1.00 g/L)(L/9.585 E-4 mol)
⇒ mm = 1043.33 g/mol
<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass.
For this, we look at the atomic masses of the elements present in the compound.
Cu has an atomic mass of 63.546 amu
Fe has 55.845 amu
and S has 36.065 amu
Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu
So we have not established the mass of the compound in amus
63.546 + 55.845 + 72.13 = 191.521
That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845
So to get the percentage, or fraction of iron, we take 55.845 / 191.521
Which comes out to 29.15% by mass
Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>